Let
$E_4(z)= - \frac{B_4}{8}+ \sum_{n=1}^\infty \sigma_3(n) q^n$
and
$E_6(z)= - \frac{B_6}{12}+ \sum_{n=1}^\infty \sigma_5(n) q^n$
How does one show they are algebraically independant over $\mathbb{C}$ ?
Let
$E_4(z)= - \frac{B_4}{8}+ \sum_{n=1}^\infty \sigma_3(n) q^n$
and
$E_6(z)= - \frac{B_6}{12}+ \sum_{n=1}^\infty \sigma_5(n) q^n$
How does one show they are algebraically independant over $\mathbb{C}$ ?